Physical Science-please help
posted by Sarah on .
A 250.0 kg roller coaster car has 20000 J of potential energy at the top of a hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?
The KE will equal the PE at the top
1/2 m v^2= 20000
solve for v.
Could you please put this in layman's terms?
I believe that this answer is 9.8 m/s am I right?
The kinetic energy (1/2 *mass*velocity squared) will equal the potential energy a the top
1/2 m v2 = 20000
The veloicity will be near 13 meters/second at the bottom